Download Modular Arithmetic - svmoore

Modular Arithmetic
Shirley Moore
CS4390/5390 Fall 2013
http://svmoore.pbworks.com/
September 3, 2013
Agenda
• Complete pre-course survey if you haven’t
already
• Discuss homework from last class (15 min)
• Homework 1 to turn in (due Sept. 12) (5 min)
• Introduction to Matlab by Rogelio Long (40
min)
• Modular arithmetic (15 min)
• Wrap-up and preparation for next class (5
min)
Technique of Proving NPCompleteness
• Prove: If B is NP-complete, C is in NP, and B is
polynomial time reducible to C, then C is NPcomplete.
• Definition: A problem L is NP-complete if and
only if:
– L is in NP
– for every problem L’ in NP, there is a polynomialtime reduction from L’ to L
• Proof:
NP-completeness Proof
• HamCyc: Given a graph G=(V,E) with at least three
vertices, does G contain a Hamiltonian cycle?
• Traveling Salesman Problem (TSP): Given n ≥ 3
cities and distances d(i,j) between each pair of
cities, does there exist a tour of length ≤ k that
visits each city and returns to the starting point?
• Given that HamCyc is NP-complete, prove that
TSP is NP-complete.
1. Show that TSP is in NP.
2. Show a polynomial time reduction from HamCyc to
TSP.
Randomized Algorithm for 3SAT
• Schoning, T., "A probabilistic algorithm for k-SAT and
constraint satisfaction problems," Symposium on
Foundations of Computer Science, 1999.
– We present a simple probabilistic algorithm for solving k-SAT
and more generally, for solving constraint satisfaction problems
(CSP). The algorithm follows a simple local search paradigm:
randomly guess an initial assignment and then, guided by those
clauses (constraints) that are not satisfied, by successively
choosing a random literal from such a clause and flipping the
corresponding bit, try to find a satisfying assignment. If no
satisfying assignment is found after O(n) steps, start over again.
Our analysis shows that for any satisfiable k-CNF-formula with n
variables this process has to be repeated only t times, on the
average, to find a satisfying assignment, where t is within a
polynomial factor of (2(1-1/k))n.
Modular Arithmetic
• The number X (mod Y) is the remainder when
X is divided by Y.
– For example: 7 (mod 3) is 1 because 7 = 2 * 3 + 1.
That is, when you divide 7 by 3, you get a
remainder of 1.
– The "modulo Y" terminology can also be used in
the following way: Z = X (mod Y), meaning that Z
and X have the same remainder when divided by
Y. For example: 7 = 25 (mod 3)because 7 = 2 * 3 +
1 and 25 = 8 * 3 + 1
Modular Multiplication
• When you take products of many numbers and you
want to find their remainder modulo n, you never need
to worry about numbers bigger than the square of n.
• Pick any two numbers x and y, and look at their
remainders (mod 7):
– a = x (mod 7)
– b = y (mod 7)
• Compare the remainder modulo 7 of the products xy
and ab:
– xy (mod 7) with ab (mod 7)
• For example, try x = 26, y = 80
Modular Multiplication of Many
Numbers
• If we want to multiply many numbers modulo
n, we can first reduce all numbers to their
remainders. Then, we can take any pair of
them, multiply and reduce again.
• For example, suppose we want to find
X = 36 * 53 * 91 * 17 * 22 (mod 29)
• What is the largest number we have to
multiply by?
Modular Exponentiation
• Suppose we would like to calculate 1143 (mod 13).
• The straightforward method would be to multiply
11 by 11, then to multiply the result by 11, and so
forth. This would require 42 multiplications.
• We can save a lot of multiplications if we do the
following:
– First write 43 as a sum of powers of 2:
43 = 32 + 8 + 2 + 1
– That means that 1143 = 1132 * 118 * 112 * 11 .
• How many multiplications are required, and what
is the largest number we have to multiply?
Preparation for Next Class
• Introduction to Modular Arithmetic and
Cryptography
• Start on Homework 1 (turn in for grade, due
September 12)